Practice Systems of Equations: Algebra 1 Unit 5.
This unit covers graphing systems, substitution method and elimination method — essential concepts for Algebra 1. Use our interactive study games to test your understanding, or review questions in traditional format below.
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This unit covers graphing systems, substitution method and elimination method — essential concepts for Algebra 1. Use our interactive study games to test your understanding, or review questions in traditional format below.
Key Concepts Breakdown
1 Graphing Systems
A system of equations is solved by finding the point where two lines intersect on a graph. Students must be able to identify whether a system has one solution (intersecting), no solution (parallel), or infinitely many solutions (same line). The solution is always written as an ordered pair (x, y).
Key Points
- One solution: lines intersect at exactly one point — different slopes
- No solution: lines are parallel — same slope, different y-intercepts
- Infinitely many solutions: lines are identical — same slope and same y-intercept
- To graph, convert equations to slope-intercept form (y = mx + b) first
Graph the system: y = 2x + 1 and y = -x + 4. What is the solution?
Graph y = 2x + 1 with slope 2 and y-intercept 1, and y = -x + 4 with slope -1 and y-intercept 4. The two lines cross at the point (1, 3), so the solution to the system is (1, 3). You can verify by substituting x = 1 into both equations and confirming both give y = 3.
2 Substitution Method
Substitution works by isolating one variable in one equation and plugging that expression into the other equation. This reduces the system to a single equation with one variable, which can then be solved. Always substitute back to find the second variable.
Key Points
- Isolate a variable that already has a coefficient of 1 when possible to avoid fractions
- Substitute the expression into the OTHER equation, not the same one
- Solve for the first variable, then substitute back to find the second
- Write the final answer as an ordered pair (x, y)
Solve the system: y = 3x − 2 and 2x + y = 13.
Since y is already isolated in the first equation, substitute (3x − 2) for y in the second equation: 2x + (3x − 2) = 13, which simplifies to 5x − 2 = 13, so 5x = 15 and x = 3. Substitute x = 3 back into y = 3x − 2 to get y = 7, giving the solution (3, 7).
3 Elimination Method
Elimination works by adding or subtracting the two equations to cancel out one variable. If no variable cancels immediately, multiply one or both equations by a constant to create opposite coefficients. The goal is to end up with one equation and one variable.
Key Points
- Add the equations when coefficients of one variable are already opposites (e.g., 3x and −3x)
- Multiply one or both equations by a constant to create opposite coefficients before adding
- After eliminating one variable, solve for the remaining variable
- Substitute back into either original equation to find the second variable
Solve the system: 3x + 2y = 16 and 5x − 2y = 8.
The y-coefficients are already opposites (+2y and −2y), so add the two equations directly: (3x + 5x) + (2y − 2y) = 16 + 8, which gives 8x = 24, so x = 3. Substitute x = 3 into the first equation: 3(3) + 2y = 16 → 9 + 2y = 16 → 2y = 7 → y = 3.5, giving the solution (3, 3.5).
Questions, answered.
What is Systems of Equations?
Systems of Equations is Unit 5 of Algebra 1, covering graphing systems, substitution method and elimination method.
How to study for Algebra 1 Unit 5?
Start with the Quick Summary above, review the Key Concepts, then test yourself with our interactive study games. Aim for 80%+ accuracy before moving on.
How many questions are in this unit?
This unit has 28+ review questions across 5 different game modes.